Maple Solutions to the Chemical Engineering Problem Set

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DEqns2:=subs(AEqns,AEqns,DEqns): > DEqns3:=subs(params,startvalues,DEqns2): DEqns3; ⎡ ⎛ 5027.664181 ⎞ ⎢ ⎜11.17258707 − ⎟ ⎢ ∂ ⎝ T ⎠ ⎢ X = .1000000000 e ⎢∂W ⎢ ⎣ ⎛ ⎞ ⎜ ( ) ⎟ ⎜ 14871.80250 − ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ − 1 X 2 y 2 ( 1 − .5 X ) 2 T 2 Xy ∂ .002439000000 T = ⎛ 4811.161896 ⎞ ∂W ⎜− 10.69147088 + ⎟ ⎝ T ⎠ ( 1 − .5 X) Te , ⎛ 5027.664181 ⎞ ⎜ 11.17258707 − ⎟ ⎝ T ⎠ 2.000000000 − .004000000000 T + 100.0000000 e ⎛ ⎞ ⎜ ( ) ⎟ ⎜ 14871.80250 − ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ − 1 X 2 y 2 ( 1 − .5 X ) 2 T 2 Xy .002439000000 , ⎛ 4811.161896 ⎞ ⎜− 10.69147088 + ⎟ ⎝ T ⎠ ( 1 − .5 X) Te ∂ ( ) y = −.00001666666667 ∂W − ⎤ ⎥ 1 .5 X T ⎥ y ⎥ ⎦ > > > IC := [W=0,X=0,T=450,y=1]: IC; [ W= 0 , X = 0 , T = 450 , y = 1 ] > result := BESIRK(DEqns3,IC,0..20,hmax=0.2): > plot({seq(makelist(result,1,k),k=[2])}); > > DAEqns := [op(DEqns),op(AEqns)]; 0.25 0.2 0.15 0.1 0.05 0 5 10 15 20 Page 40
∂ A ∂ DAEqns X = − = ∂W F ∂W A, 0 T U ( T − T) + r ΔH ( ) ⎛ ⎞ a a A R ∂ 1 α( 1 + εX) T ⎜ 2 ⎟ y = − r = −k ⎜ C − ⎟ F C ∂W 2 yT A ⎜ A ⎟ A, 0 P, A 0 ⎝ ⎠ C ⎡ ⎢ C := ⎢ , , , , ⎢ K ⎣ C ⎛ ⎛ 1 1 ⎞ ⎞ ⎛ ⎛ 1 1 ⎞ ⎞ ⎜ ⎟ ⎜ E ⎜ − ⎟ ⎜ A ⎟ ⎜ ⎜ T T ⎟ ⎜ ΔH ( ) ⎜ − ⎟ ⎟ ⎟ ⎜ ⎝ ref ⎠ ⎜ R ⎜ T T ⎟ ⎟ ⎟ ⎜ ⎝ ref ⎠ ⎟ ⎜ ⎟ ⎜ ⎟ C ( 1 − X) yT ⎝ R ⎠ ⎝ R ⎠ A, 0 0 k = k e , K = K e , C = , ref C C, ref A ( 1 + εX) T ⎤ ⎥ εC XyT ⎥ A, 0 0 ⎥ C = − ⎥ C ( 1 + εX) T ⎥ ⎦ > DAEqns2:=Subs(params,startvalues,DAEqns); ∂ DAEqns2 = ∂W X −.2000000000 r ∂ T = − − A ∂W 2.000000000 .004000000000 T 200.0000000 r ⎡ ⎢ := ⎢ , , ⎢ A ⎣ ∂ ( ) y = −.00001666666667 ∂W − ⎛ ⎞ 1 .5 X T ⎜ 2 ⎟ r = −k ⎜ C − ⎟ y A ⎜ A ⎟ ⎝ ⎠ C ⎛ 5027.664181 ⎞ ⎜ 11.17258707 − ⎟ C ⎝ T ⎠ , , k = .5 e , K C ⎛ ⎜ − 10.69147088 + ⎝ K = 25000 e C > DAEIC:=[op(IC),op(AIC)]; 4811.161896 ⎞ ⎟ T ⎠ ( ) C = 121.950 A − ⎤ ⎥ 1 X y Xy ⎥ , , C = 60.9750 ( 1 − .5 X) T C ⎥ ( 1 − .5 X) T ⎥ ⎦ DAEIC := [ W= 0 , X = 0 , T = 450 , y = 1 , r = -.0367205 , k = .5 , K = 25000 , C = .2710000000 , C = 0] A C A C > DAEresult:=BESIRK(DAEqns2,DAEIC,0..20): polation error polation error polation error polation error polation error polation error polation error polation error The concentrations are variable numbers 8 and 9 in each line of <strong>the</strong> output table; we can plot <strong>the</strong> concentration profiles as follows > plot({seq(makelist(DAEresult,1,k),k=[8,9])}); Page 41
Page 1 and 2: Maple Solutions to the Chemical Eng
Page 3 and 4: The Van der Waals Equation of State
Page 5 and 6: dimensionless form (in terms of the
Page 7 and 8: lprint(`Reduced pressure is `, Pr);
Page 9 and 10: Steady State Material Balances on a
Page 11 and 12: set of independent equations for th
Page 13 and 14: x = .54 , x = .35 , F x = F x + F x
Page 15 and 16: and a cubic is > p(3); a + a T + a
Page 17 and 18: -40 1000. 100. 10. -20 1. 20 40 60
Page 19 and 20: -40 1000. 100. 10. -20 1. 20 40 60
Page 21 and 22: ( Z, R1 ) = 0 ( D, R3) = 0 ( C, R2
Page 23 and 24: C = .9256703707 + .1402769617 I , e
Page 25 and 26: Terminal Velocity of Falling Partic
Page 27 and 28: Make a list of tank numbers > Units
Page 29 and 30: Diffusion and Reaction in a One Dim
Page 31 and 32: IC := {C[A](0)=0.2,y(0)=y0}; Alleqn
Page 33 and 34: Kvalues := [seq(RAOULT(i),i=compid)
Page 35 and 36: ead `c:/maple/numerics/newton.mpl`:
Page 37 and 38: 108 106 104 102 100 98 96 0.4 0.5 0
Page 39: ∂ 1 α( 1 + εX) T y = − ∂W 2
Page 43 and 44: 0 -0.2 -0.4 -0.6 -0.8 5 W 10 15 20
Page 45 and 46: deproc := proc(n,t,y,dy) local q,qs
Page 47 and 48: params := { V = , , , , , , , } 400

Maple Solutions to the Chemical Engineering Problem Set

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